Below are the jersey numbers of 11 players randomly selected from a football team. Find the range, variance, and standard devia
1 answer:
Answer:
Option(A) is correct
Jersey numbers are nominal data that are just replacements for names, so the resulting statistics are meaningless.
Step-by-step explanation:
The given data set in the question are ;33, 29, 97, 56, 26, 78, 83, 74, 65, 47, 58
the range can be determined by finding the highest value and subtract it to the lowest value. In this case the values are:
Highest = 97
Lowest = 71
Range = highest value - Minimum value
Range = 97 - 26 = 71
mean of the data is the summation of all the numbers in the data set divided by the number of given samples.
Mean = (33 + 29 + 97 + 56+ 26 + 78 + 83 74+ 65 + 47 + 58)/11
= 647/11
Now to find the variance of the data set by using below formular
σ²=[ (xᵢ -mean)²]/n-1
[(33-58.7)² +(29-58.7)²+( 97-58.7)²+( 56-58.7)²+( 26 -58.7)²+(78-58.7)²+( 83 -58.7)²+(74-58.7)²+( 65-58.7)²+( 47 -58.7)²+(58 -58.7)²]/10
Now, we will calculate standard deviation by taking square root over variance
σ =√(variance)
σ =√(546)
Hence, the range is 71 ,variance is 546 and standard deviation is 23.4 therefore,
Option A is the answer that is Jersey numbers are nominal data that are just replacements for names, so the resulting statistics are meaningless.
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Record of june temperature of last 50 years
Probability of June temperature being above 80° F from last 50 years
It means , out of 30 days ,there are only 20 days , in which temperature is above 80°F.
⇒Probability that it will be above 80° F for 20 or more days this June
where, x is the number of extra days after 20 days in month of june, when temperature is above 80°F.
Value of this Probability will be in between
Now, what Azul wants
To find the probability that it will be above 80° F for 20 or more days this June.
The Simulation given is
⇒There are 30 days in June. If we use a number cube with the numbers 1-6 and assign 1-4 as a temperature above 80° F and 5-6 as a temperature 80° F or below.
Total number of numbers on number cube ={1,2,3,4,5,6}
Favorable Outcome={1,2,3,4}=Temperature above 80° F
Temperature 80° F or below ={5,6}
⇒Probability of getting number {1,2,3,4} on the number cube when rolled once,that is Temperature above 80° F
⇒Probability of getting number {5,6} on the number cube when rolled once, that is Temperature 80° F or below
This is Incorrect Simulation, as there are chances that , there can be more than 20 days out of 30 days when Temperature goes above 80° F.But Here we have got Probability of temperature being above 80° F for 20 days only, not more than 20 days, which is
Not,a number
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